Multiple choice

A man has $3$ coins $A, B$ & $C$. $A$ is fair coin. $B$ is biased such that the probability of occurring head on it is $2/3$. $C$ is also biased with the probability of occurring head as $1/3$. If one coin is selected and tossed three times, giving two heads and one tail, find the probability that the chosen coin was $A$

  1. $9/25$
  2. $3/5$
  3. $27/125$
  4. $1/3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let E be the event of getting 2 heads and 1 tail in 3 tosses. P(E|A) = 3 * (1/2)^2 * (1/2) = 3/8. P(E|B) = 3 * (2/3)^2 * (1/3) = 12/27 = 4/9. P(E|C) = 3 * (1/3)^2 * (2/3) = 6/27 = 2/9. Using Bayes' Theorem, P(A|E) = (P(A)P(E|A)) / (P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C)). Assuming P(A)=P(B)=P(C)=1/3, P(A|E) = (3/8) / (3/8 + 4/9 + 2/9) = (3/8) / (3/8 + 6/9) = (3/8) / (3/8 + 2/3) = (3/8) / (25/24) = 9/25.

AI explanation

Using Bayes' theorem, the probability of selecting coin A given two heads and one tail is the probability of the outcome given A, divided by the total probability of the outcome across all coins. The probability of 2 heads and 1 tail is (1/2)^3 = 1/8 for coin A, (2/3)^2 times (1/3) = 4/27 for coin B, and (1/3)^2 times (2/3) = 2/27 for coin C. The total probability is 1/3 times (1/8 + 4/27 + 2/27), which equals 1/3 times (27/216 + 32/216 + 16/216) = 1/3 times 75/216 = 25/216. The probability that the chosen coin was A is the ratio of its part to the total, (1/3 times 1/8) divided by 25/216, which equals 9/216 divided by 25/216, giving 9/25.

Keep practicing — related questions